Question

Two points are marked on X axis and Y axis as (-3,0) and (3,0) respectively. Find (1) Distance between these two points. (ii) Slope of line passing through these points (iii) Inclination of the line passing through these two points.

Answer 1

(i)Therefore the distance between (-3,0) and (0,3) is  $=3\sqrt{2}$ unit

(ii)Therefore the slope of the required line is = 1

(iii)Therefore the line passes through the given point makes an angle 45° with the positive x-axis.

Step-by-step explanation:

Given points are (-3,0) and (0,3).

(i)  

The distance between (x₁,y₁,z₁) and (x₂,y₂,z₂) is  $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here x₁=-3, y₁ = 0 , x₂=0  and  y₂= 3

Therefore the distance between (-3,0) and (0,3) is $=\sqrt{(0+3)^2+(3-0)^2}$

                                                                                       $=3\sqrt{2}$ unit

(ii)

Slope of a line passes through the points (x₁,y₁,z₁) and (x₂,y₂,z₂) is

$\frac{y_2-y_1}{x_2-x_1}$

Therefore the slope of the required line is= $\frac{3-0}{0+3}$ = 1

(iii)

Slope (m) = tan θ

⇒tan θ = 1

⇒θ = tan⁻¹(1)

⇒θ = 45°

Therefore the line passes through the given point makes an angle 45° with the positive x-axis.

Answer 2

Two points are marked on X axis and Y axis as (-3,0) and (3,0) respectively.

1. Distance between these two points

= √ ( x1 - x2 )^2 + ( y1 - y2 )^2

= √ ( -3 -3 )^2 + ( 0 - 0 )^2

= √ ( -6 ) ^2 = √ 36 = 6

2. Slope = y2 - y1 / x2 - x1

So, slope of these two points

= 0 - 0 / 3 + 3 = 0 / 6 = 0

3. Slope = tan (theta)

tan (theta) = 0

Therefore, theta = 0°

Line through these two points makes 0° angle with x-axis